Hello All,
I am in need of your help.
I have a scenario. I have two columns, one titled HIV_Date and the other x1st_service
the dataset I am using is just called Patient_Demographics
what I want to do is in each column (HIV_Date & x1st_service) I want to filter out the year 2021 so just 2021 cases show.
I used the coding below, but I got back 0 rows
Demographic.Data %>%
filter(HIV.Date == 2021 | X1st.Service == 2021)
0 rows | 1-8 of 17 columns = output
What would the proper method be to be able to filter out the data set by this ask? I have not converted the dates by using MDY or as.date either.
All the help is greatly appreciated.
Thank you!
Hi,
In order for us to help you, it's easier if you create a reprex which we can work with. A reprex consists of the minimal code and data needed to recreate the issue/question you're having. You can find instructions how to build and share one here:
A minimal reproducible example consists of the following items:
A minimal dataset, necessary to reproduce the issue
The minimal runnable code necessary to reproduce the issue, which can be run
on the given dataset, and including the necessary information on the used packages.
Let's quickly go over each one of these with examples:
Minimal Dataset (Sample Data)
You need to provide a data frame that is small enough to be (reasonably) pasted on a post, but big enough to reproduce your issue.
Let's say, as an example, that you are working with the iris data frame
head(iris)
#> Sepal.Length Sepal.Width Petal.Length Petal.Width Species
#> 1 5.1 3.5 1.4 0.…
Normally the filtering you wrote should work like so:
library(dplyr)
set.seed(2)
myData = data.frame(
x = runif(10),
HIV.Date = sample(c(2020,2021), 10, replace = T),
X1st.Service = sample(c(2020,2021), 10, replace = T)
)
myData %>% filter(HIV.Date == 2021 | X1st.Service == 2021)
#> x HIV.Date X1st.Service
#> 1 0.5733263 2021 2020
#> 2 0.1680519 2020 2021
#> 3 0.1291590 2021 2021
#> 4 0.8334488 2021 2020
#> 5 0.4680185 2021 2021
#> 6 0.5499837 2021 2021
Created on 2021-07-13 by the reprex package (v2.0.0)
Post a reprex and we can take it from there
PJ
system
Closed
August 3, 2021, 3:54pm
3
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